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I have a question about surface of revolution.

Prove that a surface of revolution is a 2dimension manifold. enter image description here

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hint: (u,v) are local coordinates.

Edit: the standard parametrization of a surface of revolution is $(f(v)\cos(u), f(v)\sin(u),g(v))$, where the given curve was $(f(v), g(v))$. Now, try to prove that locally (u,v) are coordinates (in few words, (u,v) give the local euclidean structure to the surface). Hope this helps.

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    $\begingroup$ So? Please write more clear:/ @user67133 $\endgroup$
    – 1190
    Commented Mar 20, 2013 at 19:37
  • $\begingroup$ Unless there is a very good reason, we prefer not to arbitrarily delete good content: the answers could well help other users in the future. $\endgroup$
    – robjohn
    Commented May 13, 2013 at 1:06

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